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# Find n-th term in series 1 2 2 3 3 3 4 4 4 4….

• Difficulty Level : Medium
• Last Updated : 07 Jun, 2022

Given series 1 2 2 3 3 3 4 4 4 4 …., find n-th term of the series. The “pattern” is obvious. There is one “1”, two “2”s three “3”s, etc.

Examples:

```Input : n = 5
Output : 3

Input :  n = 7
Output : 4```

A naive approach is to run two loops one from 1 to n and the other from 1 to i, and for every iteration of the inner loop keep a count, whenever the count reaches n, we can break out of both the loops and i will be our answer.
Time Complexity: O(n), as we will be using a loop to traverse n times.

Auxiliary Space: O(1), as we will not be using any extra space.

An efficient approach will be to note down a small observation which is:
The trick is to find a pattern.
Consider numbering the given sequence as follows:
1 is at position 1
2 is at positions 2, 3
3 is at positions 4, 5, 6
4 is at positions 7, 8, 9, 10
and so on…
Notice that the last positions of the individual values form a sequence.
1, 3, 6, 10, 15, 21…
If we want a formula for the “nth” term, start by looking at it the other way around. In which term does the number “n” first appear? Taking the first term to be the “0th” term. “1” appears in term 0, “2” appears in term 1, “3” appears in term 1+2=3, “4” appears in term 1+2+3= 6, etc. The number “x” first appears in term 1 + 2 + …+ (x- 2)+ (x-1) = x(x-1)/2.
So solving for the n-th term we get n = x*(x-1)/2
Solving it using quadratic equation we get

x = ( ( 1 + sqrt(1+8*n) )/2 )

n in every case is NOT an integer which means the n-th number is not the first of a sequence of the same integer, but it is clear that the n-th integer is the integer value.

## C++

 `// CPP program to find the nth term of the series` `// 1 2 2 3 3 3 ...` `#include ` `using` `namespace` `std;`   `// function to solve the quadratic equation` `int` `term(``int` `n)` `{` `    ``// calculating the Nth term` `    ``int` `x = (((1) + (``double``)``sqrt``(1 + (8 * n))) / 2);` `    ``return` `x;` `}`   `// driver code to check the above function` `int` `main()` `{` `    ``int` `n = 5;` `    ``cout << term(n);` `    ``return` `0;` `}`

## Java

 `// Java program to find the nth` `// term of the series 1 2 2 3 3 3 ...` `import` `java.io.*;`   `class` `Series {` `    `  `    ``// function to solve the quadratic` `    ``// equation` `    ``static` `int` `term(``int` `n)` `    ``{` `        ``// calculating the Nth term` `        ``int` `x = (((``1``) + (``int``)Math.sqrt(``1` `+ ` `                           ``(``8` `* n))) / ``2``);` `        ``return` `x;` `    ``}` `    `  `    ``// driver code to check the above function` `    ``public` `static` `void` `main (String[] args) {` `        ``int` `n = ``5``;` `        ``System.out.println(term(n));` `    ``}` `}`   `// This code is contributed by Chinmoy Lenka`

## Python3

 `# Python program to find the nth term ` `# of the series 1 2 2 3 3 3 ...` `import` `math `   `# function to solve the quadratic equation` `def` `term( n ):`   `    ``# calculating the Nth term` `    ``x ``=` `(((``1``) ``+` `math.sqrt(``1` `+` `(``8` `*` `n))) ``/` `2``)` `    ``return` `x`   `# Driver code` `n ``=` `5` `print``(``int``(term(n)))`   `# This code is contributed by Sharad_Bhardwaj.`

## C#

 `// C# program to find the nth` `// term of the series 1 2 2 3 3 3 ...` `using` `System;`   `class` `Series` `{` `    ``// function to solve the quadratic` `    ``// equation` `    ``static` `int` `term(``int` `n)` `    ``{` `        ``// calculating the Nth term` `        ``int` `x = (((1) + (``int``)Math.Sqrt(1 + (8 * n))) / 2);` `        ``return` `x;` `    ``}`   `    ``// driver code to check the above function` `    ``public` `static` `void` `Main()` `    ``{` `        ``int` `n = 5;` `        ``Console.WriteLine(term(n));` `    ``}` `}`   `// This code is contributed by vt_m.`

## PHP

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## Javascript

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Output:

`3`

Time Complexity: O(log(n)) as sqrt function takes O(log n).
Auxiliary Space: O(1), as we are not using any extra space.

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